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\docbegin


\section{Metric Spaces}
\begin{Exec}\label{Exec1.1}
  Let $b\neq 0$. Prove that $|a+b|=|a|+|b|$
  if and only if there exists a real constant $k \geq 0$ such that $a= kb$.
\end{Exec}
\begin{proof}
  Since $|a+b|^2 = (a+b)\overline{(a+b)} =
  |a|^2 + |b|^2 + 2\Re{(\overline{a}b)} \leq |a|^2+|b|^2 + 2|a||b|$,
  it follows that $|a+b|=|a|+|b|$
  if and only if $\Re{(\overline{a}b)} = |a||b|$.

  Since $b \neq 0$, let $k=a/b$. It gives that $|k|=|a|/|b|$.
  So $\Re{(k)} = \Re{(\overline{a}b)}/|b|^2 = |a|/|b|$.
  Thus $k =\Re (k)= |a|/|b| \geq 0$.
\end{proof}

\begin{Exec}
  Let $p\geq 1$.  Prove that $\big(\sum\limits_{k=1}^n|a_k|\big)^p\leq
  n^{p-1}\sum\limits_{k=1}^n|a_k|^p$.
\end{Exec}

\begin{proof} Clearly, the inequality holds for $p=1$. If $p>1$ then by the H\"older  inequality,
  \[
  \sum\limits_{k=1}^n |a_k| \leq
  \left(\sum\limits_{k=1}^n 1^{\frac{p}{p-1}}\right)^{1-\frac{1}{p}}
  \left(\sum\limits_{k=1}^n |a_k|^p\right)^{\frac{1}{p}}
  = n^{\frac{p-1}{p}} \left(\sum\limits_{k=1}^n |a_k|^p\right)^{\frac{1}{p}}.
  \]
  Thus $\big(\sum\limits_{k=1}^n|a_k|\big)^p\leq
  n^{p-1}\sum\limits_{k=1}^n|a_k|^p$ holds for $p\geq 1$.
\end{proof}

\begin{Exec}
  In a metric space $(X,d)$ prove that
  \[
  |d(x,y)-d(x',y')| \leq d(x,x') + d(y,y') .
  \]
  Deduce that $d(x_n,x) \rightarrow 0$ and
  $d(y_n,y) \rightarrow 0$ as $n \rightarrow \infty$ imply
  \[
  d(x_n,y_n) \rightarrow d(x,y) \quad (n \rightarrow \infty) .
  \]
\end{Exec}
\begin{proof}
  By the triangle inequality, $d(x,y) \leq d(x,y') + d(y,y')$
  and $d(x,y') \leq d(x',y')+ d(x,x')$.
  So $d(x,y) - d(x',y') \leq d(x,x')+d(y,y')$. By the same way,
  we can get $d(x',y') - d(x,y) \leq d(x,x')+d(y,y')$.
  Thus $|d(x,y)-d(x',y')| \leq d(x,x') + d(y,y')$.

  Since $|d(x_n,y_n) - d(x,y)| \leq d(x_n,x) + d(y_n,y)$,
  it follows that
  \[
  \limsup_{n\rightarrow \infty} |d(x_n,y_n) - d(x,y)|
  \leq \lim_{n\rightarrow \infty} [d(x_n,x) + d(y_n,y)] = 0.
  \]
  Then $d(x_n,y_n) \rightarrow d(x,y)$($n \rightarrow \infty$)
  follows immediately.
\end{proof}

\begin{Exec}
  Suppose that $(X,d)$ is a metric space.
  Prove that each of the following functions $\rho$ is a metric on the set $X$:
  \begin{alphlist}
  \item For every $x,y \in X$, $\rho(x,y) = \min \{1, d(x,y)\}$.
  \item For every $x,y \in X$, $\rho(x,y) = \frac{d(x,y)}{1+d(x,y)}$.
  \end{alphlist}
\end{Exec}

\begin{proof} (a)
  It is easy to check that (M1)-(M2) are satisfied.
  For every $x,y,z \in X$,
  to check (M3), notice that if $d(x,z) \geq 1 $ or $d(z,y) \geq 1$
  then
  \[
  \begin{split}
    \rho(x,y) &= \min\{1, d(x,y)\}
    \leq \min\{1, d(x,z)+d(z,y)\} = 1 \\
    &\leq \min\{1, d(x,z)\} + \min\{1, d(y,x)\}
    = \rho(x,z) + \rho(z,y) .
  \end{split}
  \]
  One the other hand, if $d(x,z) < 1 $ and $d(z,y) < 1$,
  then
  \[
  \begin{split}
    \rho(x,y) &= \min\{1, d(x,y)\}
    \leq \min\{1, d(x,z)+d(z,y)\} \leq d(x,z)+d(z,y) \\
    &= \min\{1, d(x,z)\} + \min\{1, d(y,x)\}
    = \rho(x,z) + \rho(z,y)
  \end{split}
  \]

  Hence $(X,\rho)$ is a metric space.

(b)
  (M1)-(M2) are trivial.
  For every $x,y,z \in X$,
  to check (M3), notice that
  \[
  \begin{split}
    & \rho(x,z) + \rho(z,y) - \rho(x,y) =
    \frac{d(x,z)}{1+d(x,z)} + \frac{d(z,y)}{1+d(z,y)}
    -  \frac{d(x,y)}{1+d(x,y)}  \\
    & = \frac{d(x, y) d(x, z) d(y, z) + 2 d(x, z) d(y, z)
      + [d(y, z) + d(x, z)  - d(x, y)]
    }{(1+d(x,z))(1+d(z,y))(1+d(x,y))} \geq 0.
  \end{split}
  \]

  Hence $(X,\rho)$ is a metric space.
\end{proof}

\begin{Exec}
  Prove that the space $s$ of Example 1.2.5 is a metric space. Show that
  \[
  \sup d(x,y)=1,
  \]
  where the supremun is taken over all elements x,y in $s$.
  (Geometrically this says that the space s has diameter 1.)
\end{Exec}

\begin{proof}  To show $s$ is a metric space we need only to check
(M3).
  Let $x,y,z \in s$ and $x = \{\xi_k\},y = \{\eta_k\},z = \{\mu_k\} $.
  By the triangle inequality, we have
  \[
  \begin{split}
    d(x,y) &= \sum_{k=1}^{\infty} \frac{1}{2^k}
    \frac{|\xi_k-\eta_k|}{1+|\xi_k-\eta_k|}
    \leq  \sum_{k=1}^{\infty} \frac{1}{2^k}
    (\frac{|\xi_k-\mu_k|}{1+|\xi_k-\mu_k|}+
    \frac{|\mu_k-\eta_k|}{1+|\mu_k-\eta_k|}) \\
    &= \sum_{k=1}^{\infty} \frac{1}{2^k}\frac{|\xi_k-\mu_k|}{1+|\xi_k-\mu_k|}
    + \sum_{k=1}^{\infty} \frac{1}{2^k}\frac{|\mu_k-\eta_k|}{1+|\mu_k-\eta_k|}
    = d(x,z)+d(z,y)
  \end{split}
  \]
for all series are convergent. Hence $(X,\rho)$ is a metric space.

  Since $|\xi_k-\eta_k|/(1+|\xi_k-\eta_k|) < 1$ for each $k\in\bz$,
  it gives that
  \[
  d(x,y) = \sum_{k=1}^{\infty} \frac{1}{2^k}\frac{|\xi_k-\eta_k|}{1+|\xi_k-\eta_k|}
  \leq  \sum_{k=1}^{\infty} \frac{1}{2^k} = 1 .
  \]

  Let $x_n = (n, n , \ldots)$, $y = (0,0,\ldots)$.
  So
  \[
  \lim_{n\rightarrow \infty} d(x_n,y)
  = \lim_{n\rightarrow \infty} \sum_{k=1}^{\infty} \frac{1}{2^k} \frac{n}{1+n}
  = \lim_{n\rightarrow \infty} \frac{n}{1+n} = 1 .
  \]
  Thus $\sup d(x,y)=1$.
\end{proof}

\begin{Exec}
  Let $(X_1, d_1)$ , $(X_2, d_2)$ be metric space and let
  $x=(x_1, x_2)$,$y=(y_1, y_2)$ be in the product set $X_1 \times X_2$.
  Define $d(x,y) = \max\{d_1(x_1,y_1), d_2(x_2,y_2)\}$.
  Find whether $(X_1\times X_2, d)$ is a metric space.
\end{Exec}
\begin{proof}
  Let $x,y,z \in X_1 \times X_2$ and
  $x=(x_1, x_2)$,$y=(y_1, y_2)$, $z=(z_1, z_2)$.
  By the definition, $d(x,y)=0$ if and only if $d(x_1,y_1)=0, d(x_2,y_2)=0$.
  So $x=y$ and (M1) is satisfied.
  It's easy to check that (M2) is satisfied.
  \[
  \begin{split}
    d(x,y) &= \max\{d_1(x_1,y_1), d_2(x_2,y_2)\}
    \leq \max\{d_1(x_1,z_1)+d_1(z_1,y_1), d_2(x_2,z_2)+d_2(z_2,y_2)\} \\
    &\leq \max
    \left\{d_1(x_1,z_1)+\max\{d_1(z_1,y_1), d_2(z_2,y_2)\},
      d_2(x_2,z_2)+\max\{d_1(z_1,y_1), d_2(z_2,y_2)\}
    \right\} \\
    &= \max\{d_1(x_1,z_1), d_2(x_2,z_2)\}+ \max\{d_1(z_1,y_1), d_2(z_2,y_2)\}
    = d(x,z) + d(x,y),
  \end{split}
  \]
  which gives (M3) holds then the conclusion follows.
\end{proof}

\begin{Exec}
  Prove the inequalities stated in Remark 1.2.2.
\end{Exec}
\begin{proof}
  Let $x=\{x_k\}, y=\{y_k\} \in \br^n$.

  It's easy to check $d_1(x,y) \leq d(x,y)$ and $d_1(x,y) \leq d_2(x,y)$.
  Since
  \[
  \sqrt{\sum_{k=1}^{n}(x_k - y_k)^2} \leq
  \sqrt{\sum_{k=1}^{n} \max_{1\leq i \leq n}\{(x_i - y_i)^2\}}
  = \sqrt{n} \max_{1\leq i \leq n}\{(x_i - y_i)^2\} ,
  \]
  $d(x,y) \leq \sqrt{n}\, d_1(x,y)$.

  Since
  \[
  \sum_{k=1}^{n} |x_k-y_k| \leq \sum_{k=1}^{n} \max_{1\leq i \leq n}\{(x_i - y_i)^2\}
  = n \max_{1\leq i \leq n} \{(x_i - y_i)^2\} ,
  \]
  it gives that $d_2(x,y) \leq n d_1(x,y)$.
  Notice that
  \[
  \begin{split}
    \sum_{k=1}^{n} (x_k-y_k)^2
    &= \sum_{k=1}^{n} |x_k-y_k|^2 \\
    \leq \sum_{i=1}^{n} \sum_{j=1}^{n} |x_i-y_i||x_j-y_j|
    &= (\sum_{k=1}^{n} |x_k-y_k|)^2
  \end{split}
  \].
  So $d(x,y) \leq d_2(x,y)$.

  By the H\"older  inequallity, we get
  \[
  \begin{split}
    d_2(x,y) &= \sum_{k=1}^{n} |x_k-y_k|
    = \sum_{k=1}^{n} |x_k-y_k|\times 1 \\
    &\leq \sqrt{\sum_{k=1}^{n} |x_k-y_k|^2} \sqrt{\sum_{k=1}^{n} 1^2}
    = \sqrt{n} d(x,y)
  \end{split}
  \].
\end{proof}

\begin{Exec}
  In a metric space $(X,d)$ prove that the sequence $\{x_n\}$
  converges if and only if every subsequence of $\{x_n\}$ converges.
\end{Exec}
\begin{proof} ($\Rightarrow$)\quad
Let $\{x_{n_k}\}$
  be the subsequence of $\{x_n\}$ and
  $x_n \rightarrow x \in X$.
  Since $\lim_{n\rightarrow 0} d(x_n,x) = 0$,
  it gives that $\lim_{k\rightarrow 0} d(x_{n_k},x) = 0$.
  Hence $x_{n_k} \rightarrow x$.



  ($\Leftarrow$)\quad Suppose that every subsequence of $\{x_n\}$ converges.
  Since $\{x_n\}$ is the subsequence of $\{x_n\}$,
  it follows that $\{x_n\}$ converges.
\end{proof}

\begin{Exec}
  In a metric space $(X,d)$ prove that if a Cauchy sequence has
  a convergent subsequence then the whole sequence is convergent.
\end{Exec}

\begin{proof}
  Let $\{x_{n_k}\}$ be the convergent subsequence of
  a Cauchy sequence $\{x_n\}$ such that
  $x_{n_k} \rightarrow x_0$ as $k \rightarrow \infty$.   Then
for every $\varepsilon > 0 $, there exists $K \in \bn$ such that
  \[
  d(x_{n_K}, x_0) < \frac{\varepsilon}{2}.
  \]
  Since $\{x_n\}$ is a Cauchy sequence,
  there exists $M \in \bn$ such that
  \[
  d(x_{n}, x_m) < \frac{\varepsilon}{2} \quad \text{for all}\ n,m > M .
  \]
  Taking $N=\max\{n_K, M\}$, we obtain from the above that
  \[
  d(x_{n},x_0) \leq d(x_{n},x_{n_K}) + d(x_{n_K},x_0) < \varepsilon
  \quad \text{for all } n > N.
  \]
  Thus the whole sequence $\{x_{n_k}\}$  is convergent.
\end{proof}


\begin{Exec}
\label{Exec1.10}
  \begin{alphlist}
    \item  If $A\subset B$ then $A'\subset B'$, $A^\circ\subset
    B^\circ$ and $\overline{A}\subset \overline{B}$.
    \item $(A\cup B)'=A'\cup B'$.
    \item $A$ is open if and only if $A=A^\circ$.
    \item $A$ is closed if and only if $A=\overline{A}$.
  \end{alphlist}
\end{Exec}

\begin{proof}
  \begin{alphlist}
  \item Suppose that  $x \in A'$. For every $\varepsilon>0$,
    the open ball $B(x_0, \varepsilon)$ contains infinitely many
    points of A. Since $A \subset B$, it gives that
    $B(x_0,\varepsilon)$ contains infinitely many points of B.
    Hence $x \in B'$, so that $A' \subset B'$.

    If $x \in A^\circ$, there exists $r>0$ such that the open ball
    $B(x,r)\subset A$. Take $A \subset B$. It follows that
    $B(x,r) \in B$, so that $A^\circ \subset B^\circ$.


    Since $\overline{A} = A' \cup A$ and
    $\overline{B} = B' \cup B$, it follows that
    $\overline{A} \subset \overline{B}$


  \item $A'\cup B' \subset (A\cup B)'$ follows from (a).
    For every $ x \in (A\cup B)'$, there exists a
    sequence $\{x_n\}$ in $A\cup B$ such that $x_n \neq x $
    and $x_n \rightarrow x$ as $n \rightarrow \infty$.
    So at least one of $\{x_n\} \cap A$ and  $\{x_n\} \cap B$
    is an infinite set.

    Suppose that $\{x_n\} \cap A$ is an infinite set. Then there
    is a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $\{x_{n_i}\} \in A$
    and $x_{n_i} \rightarrow x$ as $i \rightarrow \infty$.
    It implys  that $x \in A'$. If $\{x_n\} \cap B$ is an infinite set,
    we gets $x \in B'$ in the same way.
    Hence $(A\cup B)' \subset A' \cup B'$ so that
    $(A\cup B)' \subset A' \cup B'$.
  \item If $A$ is open, every point of $A$ is an interior point
    of A. So $A \subset A^\circ$.
    Since $A^\circ \subset A$, it follows that
    $A=A^\circ$.

    Suppose that $A=A^\circ$. Since $A^\circ$ is open, $A$ is open.
  \item
    If $A$ is closed, $A$ contiains all its accumulation points.
    So $A' \subset A$. Since $\overline{A} = A \cup A'$,
    $\overline{A} \subset A$. Clearly $A \subset \overline{A}$.
    Hence $A=\overline{A}$.

    Suppose that $A=\overline{A}$. Since $\overline{A}$ is closed,
    $\overline{A}$ is closed.
  \end{alphlist}
\end{proof}

\begin{Exec}
  Consider the metric space $(X, d)$, where
  $X = [0, 3)\cup [4, 5]\cup (6, 7)\cup\{8\}$ and $d$
  is the Euclidean metric
  in $\br$ restricted to $X$. For each of the following subsets,
  indicate whether it is open and whether it is closed, and justify
  your assertions:

  \begin{tabular}{rllllllll}
    a)& $[0, 3)$ & b)& $[4, 5)$ & c)& $(6, 7)$ & d)& $\{8\}$\\
    e)& $[0,3)\cup[4, 5)$ & f)& $[0, 3)\cup (6, 7)$ & g)& $(6, 7)\cup
    \{8\}$ & h)& $[1, 2)$\\
    i)& $(1, 2)$ & j)& $[1, 2]$.
  \end{tabular}
\end{Exec}

\begin{solution}
  Since $X $ is a subspace of $\mathbb{R}$,
  by Theorem 1.4.7 and the Theorem 1.4.8, we gets
  \begin{alphlist}
  \item $[0,3)$ is closed and open.
  \item $[4,5)$ is open.
  \item $(6,7)$ is closed and open.
  \item $\{8\}$ is closed and open.
  \item $[0,3)\cup[4,5)$ is open.
  \item $[0,3) \cup (6,7) = $ is closed and open.
  \item $(6,7) \cup \{8\}$ is closed and open.
  \item $[1,2)$ is not close or opened.
  \item $(1,2)$ is open.
  \item $[1,2]$ is closed.
  \end{alphlist}
\end{solution}

\begin{Exec}
  Can an open ball with radius $4$ in a metric space be a
  proper subset of another open ball with radius $3$?
\end{Exec}
\begin{solution} Yes can. For example,
  consider the metric space $(X, d)$, where $X=\{0, 1, 2, 3, 4\}$
  and $d$ is the Euclidean metric in $\br$ restricted to X.

  Since $B(4,4) = \{1, 2, 3, 4\} $ and $B(2,3) = \{0, 1, 2, 3, 4\}$,
  it gives that
  $B(4,4) \subsetneq B(2,3)$.
\end{solution}

\begin{Exec}
  Prove the relations in Remark~\ref{Rek1.4.7}, i.e.
  $A^\circ=\bigcup_{\substack{G\subset A\\ G\,\text{open}}} G$  and
  $\overline{A}=\bigcap_{\substack{F\supset A\\
      F\,\text{closed}}}F $, where $A\subset(X, d)$.
\end{Exec}
\begin{proof}
  Since $A^\circ$ is open, it follows that $A^\circ \subset
  \bigcup_{\substack{G\subset A\\ G\,\text{open}}} G$. On the other hand,
  If  $G$ is open such that $G\subset A$ then $G = G^\circ$ and so
 $G^\circ \subset A^\circ$  by Exercises \ref{Exec1.10}.
  Hence $G \subset A^\circ$ so that
  $A^\circ=\bigcup_{\substack{G\subset A\\ G\,\text{open}}} G$.

  Since $\overline{A}$ is closed and $\overline{A} \supset A$,
  it gives that $\overline{A} \supset \bigcap_{\substack{F\supset A\\
      F\,\text{closed}}}F$.
  If $F$ is closed such that $F\supset A$ then $F = \overline{F}$ and
 $\overline{F} \supset \overline{A}$  by Exercise \ref{Exec1.10}.
  Hence $F \supset \overline{F}$ so that
  $\overline{A}=\bigcap_{\substack{F\supset A\\ F\,\text{closed}}}F $
\end{proof}

\begin{Exec}
  Let $B\subset [a,b]$. Prove that
\begin{itemize}
    \item[\ding {172}] the subset $A=\{f\in C[a,b]:f(t)=0\ \text{when}\ t\in
B\}$ of $C[a,b]$ is a closed in $C[a,b]$;
    \item[\ding {173}] the set $A=\{f\in C[a,b]:|f(t)|<a\
\text{when}\ t\in B\}\ (a>0)$ is open in $C[a,b]$ if and only if $B$
is closed in $[a,b]$.
\end{itemize}
\end{Exec}

\begin{proof}
  \begin{itemize}
    \item[\ding {172}] For every $f \in A'$, there exists a sequence
      $\{ f_n \}$ in A such that $f_n \neq f$ and $f_n \rightarrow f$
      as $n \rightarrow \infty$. So
      \[
      \sup_{t \in B} |f(t)| = \sup_{t \in B} |f(t)-f_n(t)|
      \leq \sup_{t \in [a,b]} |f(t)-f_n(t)| = d(f,f_n)
      \] for each $n$. Then we get $\sup_{t \in B} |f(t)|=0$  by letting $n\rightarrow \infty$ in the above.
      Hence $f \in A$.
    \item[\ding {173}] ($\Leftarrow$)\quad Suppose that $B$ is closed.  Let $f_0 \in A$, we have $f_0 \in C[a,b]$ and
      $|f_0| \in C[a,b]$.
      So there exists a real number $t_1 \in B$ such that
      $\max_{t \in B} |f_0(t)| = |f_0(t_1)| < a$.
     Taking $r = (a - |f_0(t_1)|)/2$, we get that for every $f \in B(f_0,r)$,
      $$
      \sup_{t\in B} |f(t)| \leq \sup_{t\in B} [|f(t)-f_0(t)| + |f_0(t)|] < a.
      $$
      Hence $B(f_0, r) \subset A$ so that $A$ is open.

      ($\Rightarrow$)\quad  We argue by contradiction.  Let $\{t_n\}\subset B$ such that $\lim_{n\to\infty }t_n=t_0$. If $t_0\not\in B$.
      Consider the function $f_0(t) = a-|t-t_0|$, then $f_0(t)=a-|t-t_0|<a$ for each $t\in B$ so that $f\in A$.
      Since $A$ is open, there exists $\delta>0$ such that  $f\in A$ for
      all $f\in C[a,b]$
      with $d(f, f_0)<\delta$.  Define $f_n(t)=a-|t-t_0|+|t_n-t_0|$
      ($n=1,2,\cdots$), then $d(f_n,f_0)=|t_n-t_0|\to 0$ as
      $n\to\infty$, it follows that $f_n\in A$ whenever
      $|t_n-t_0|<\delta$.  Note that
      $f_n(t_n)=a-|t_n-t_0|+|t_n-t_0|=a$, which contradicts with the
      definition of $f_n\in A$. Hence $t_0\in B$ so that
      $B$ is  open.

      Thus  the set $A=\{f\in C[a,b]:|f(t)|<a\
      \text{when}\ t\in B\}\ (a>0)$ is open in $C[a,b]$ if and only if $B$
      is closed in $[a,b]$.
\end{itemize}
\end{proof}

\begin{Exec}\label{Exec1.15}
  If a metric space $(X,d)$ is separable, prove that every subspace of
  $X$ is separable.
\end{Exec}

\begin{proof}
  Suppose $A$ is a subspace of $X$.
  Let $D= \{x_n : n\in \bn\}$ is a countable dense subset of $X$
  (if $D$ is finite, let $x_s = x_{s+1} = \dots$).
  We construct a countable dense subset of $A$ in the following way:
  for every $m \in \bz^+$, we choose a point $x_{m,i}$ in $B(x_i, 1/m) \cap A$
  (if $B(x_i, 1/m) \cap A = \emptyset$ , ignore it) and let
   $E_m = \{x_{m,i} : i \in \bn \}$.  Clearly $E_m$ is countable and so is the set $E = \cup_{m\in \bz^+}E_m$.

  We claim that $E$ is the dense subset of $A$.  In fact, for every $\varepsilon > 0$ and
  $x \in A$
  considering $M = [2/\varepsilon]+2$.
  It follows from
  Remark 1.4.9 that $x$  must belong to some ball of the set
  $\{ B(x_i, 1/M) : i \in \bn \}$. Let $x \in B(x_i, 1/M)$.
  Therefore
  there exists a point $y \in E$ such that $y \in B(x_i, 1/M)$.   It follows
  the construction of $E$ that $d(x,y) < 1/M + 1/M < \varepsilon$.
  Hence $A$ is separable.

\end{proof}

\begin{Exec}\label{Exec1.16}
  Suppose that $B[a,b]$ is the set of all real bounded
  functions on $[a,b]$, define a function as
  $$
  d(f,g)=\sup_{a\leq t\leq b}|f(t)-g(t)|
  $$
  for every $f, g\in B[a,b]$. Prove that this $d$ is a metric of
  $B[a,b]$ and  $B[a,b]$ with this metric is not separable.
\end{Exec}

\begin{proof} That $d$ is a metric of
  $B[a,b]$ is clear.
  If $B[a,b]$ is separable, then there exists a countable dense subset $A$
  of $B[a,b]$. Since $A$ is countable, the set
  $\{ B(x, 1/3) : x \in A \}$ is countable. It follows from Remark \ref{Rek1.4.9} that each element of $B[a,b]$ must belong to some ball of the above form.
  For each $t_0 \in [a,b]$ , consider
  \[
  f_{t_0}(t) =
  \begin{cases}
    0 & \text{if } a \leq t \leq t_0, \\
    1 & \text{if } t_0 < t \leq b,
  \end{cases}
  \]
  then $f_{t_0} \in B[a,b]$. Let $E = \{f_{t_0} : t_0 \in [a,b]\}$.
  Note that $E$ is uncountable, so there exists at least two points
  $x, y \in E$ , $x \neq y$, such that $x, y \in B(x_0, 1/3)$ for some $x_0$,
  therefore
  \[
  1 = d(x,y) \leq d(x,x_0) + d(y, x_0) \leq \frac{2}{3}
  \]
  a contradiction. So $B[a,b]$ is not separable.
\end{proof}

\begin{Exec}
  Prove that each Cauchy sequence of a metric space is bounded.
\end{Exec}
\begin{proof}
  Suppose that $\{x_n\}$ is a Cauchy sequence. So there exists
  a numeber $N >0$ such that $d(x_n, x_m) < 1$ for each $n, m > N$.
  Let $M = 1 + \max_{0 < i \leq N}{d(x_{N+1}, x_1)}$.
  Clearly, $\{x_n\} \in B(X_{N+1}, M)$. Thus $\{x_n\}$ is bounded.
\end{proof}

\begin{Exec}\label{Exec1.18}
  Prove that the spaces $c$, $c_0$ are complete.
\end{Exec}

\begin{proof}
  By Example  \ref{Exam1.5.3},  we knew that $l^\infty$ is complete.
  Clearly $c, c_0$ are the subspaces of $l^\infty$.
 If we show that $c, c_0$ are closed  in $l^\infty$ then $c, c_0$ are complete as a consequence
  of the Theorem \ref{Thm1.5.1}.  We now prove that $c, c_0$ are closed  in
  $l^\infty$.

  In fact, suppose that $x_0 = \{\xi_{0,i}\}$ is
  an accumulation point of $c$ in $l^\infty$.
  There exists a sequence $\{x_n = \{\xi_{n,i}\}\}$ in $c$ such that $x_n \neq x_0$ and
  $x_n \rightarrow x_0$ as $n \rightarrow \infty$.
  For every $\varepsilon > 0$, there exists $ N >0$ such that
  $|\xi_{N,i} - \xi_{0,i}| \leq \sup_{i\in \bn}|\xi_{N,i} - \xi_{0,i}| = d(x_N,x_0)
  < \varepsilon$ for each $i \in \bn$.
  Since $x_N$ is converge, there exists $M >0$ such that
  $|\xi_{N,i} - \xi_{N,j}| < \varepsilon$ for each $i, j > M$.
  So $|\xi_{0,i} - \xi_{0,j}| \leq |\xi_{N,i} - \xi_{0,i}| + |\xi_{N,j} - \xi_{0,j}|
  + |\xi_{N,i} - \xi_{N,j}| \leq 3 \varepsilon$ for each $i, j > M$.
  Hence $\{\xi_{0,i}\}$ is converge so that $x_0\in c$.  Hence $c$ is closed.

  Suppose that $x_0 $ is an accumulation point of $c_0$
  in  $l^\infty$. In the same way like above, there exists $N>0$ such that
  $|\xi_{N,i} - \xi_{0,i}| < \varepsilon$ for each $i \in \bn$,
  then there exists $M >0 $ such that
  $|\xi_{N,i} - \xi_{N,j}| < \varepsilon$ for each $i, j > M$.
  So $|\xi_{0,i}| \leq |\xi_{N,i}| + |\xi_{0,i} - \xi_{N,i}|  2\varepsilon$.
  Hence $x_0\in c_0$  so that $c_0$ is closed.
\end{proof}

\begin{Exec}
  The standard metric for $\ell_1=\{x=\{x_n\}:\sum_{n=1}^\infty
  |x_n|<\infty\}$ is
  $$
  d(x,y)=\sum_{n=1}^\infty |x_n-y_n|.
  $$
  Since $\ell^1\subset\ell^\infty$ we could put on the `nonstandard'
  metric
  $$
  \rho(x,y)=\sup_{n\in\bn}|x_n-y_n|.
  $$
  Prove that $(\ell^1,d)$ is complete but that $(\ell^1,\rho)$ is
  incomplete.
\end{Exec}

\begin{proof}
  Suppose that $\{x_n\}$ is a Cauchy sequence of $l^1$,
  where $x_n = \{\xi_k^{(n)} \}$ such that
  $\sum_{k=1}^{\infty} |\xi_{k}^{(n)} | < \infty$.
  Now for each $\varepsilon > 0$ there exists $N \in \bn$ such that
  $|\xi_i^m - \xi_i^n| < \varepsilon $ for all $i \in \bn$ , for all $n,m > N$.
  Hence for each $i$ the sequence of number
  $\{ \xi_i^{(1)}, \xi_i^{(2)}, \dots, \xi_i^{(n)}, \dots \}$ is a Cauchy sequence
  of $\br$ or $\bc$, whence by the completeness of $\br$ or $\bc$,
  it converges to, say, $\xi_i$.  Let $x = \{\xi_i\}$.
  For $\varepsilon = 1$, there exists $N > 0$ such that
  $\sum_{i=0}^{\infty} |\xi_i^{(n)} - \xi_i| \leq \varepsilon $.
  Therefor $\sum_{i=0}^{\infty} |\xi_i|
  \leq \sum_{i=0}^{\infty} ( |\xi_i^{(N)} - \xi_i| + |\xi_i^{(N)}|)
  \leq \sum_{i=0}^{\infty} |\xi_i^{(N)} - \xi_i| + \sum_{i=0}^{\infty}|\xi_i^{(N)}|
  < \infty$,
  it implies that $x \in l^1$ so that $(l^1, d)$ is complete.

  $(l^1, \rho)$ is a subspace of $l^\infty$.
  Consider the sequence $\{x_n = \{\xi_i^{(n)}\}\}$ such that
  \[
  \xi_i^{(n)} =
  \begin{cases}
    \frac{1}{i} & \text{if } i \leq n, \\
    0 & \text{if } i > n.
  \end{cases}
  \]
  Clearly $\{x_n\}$ is a Cauchy sequence in  $(l^1, \rho)$
  and $x_n \rightarrow x_0=\{1, 1/2, \dots, 1/n, \dots \}$ in $l^\infty$.
  Since $\sum_{i=0}^{n}1/i$ is unbound, it follows that
  $\{x_n\}$ is not convergent. Hence $l^1$
  is not closed in $l^\infty$ so that $(l^1, \rho)$ is incomplete.
\end{proof}

\begin{Exec}
  Let $X$  be the set of all continuous functions on
  [a,b]. Define a metric on $X$ by
  $$
  \rho(x,y)=\int_a^b |x(t)-y(t)|\, dt,
  $$
  Then the metric space $(X, \rho)$ (not $C[a,b]$) is incomplete.
\end{Exec}

\begin{proof}
We claim that the sequence $x_n$ of $(X,\rho)$ defined by
 \[
 x_n(t)=\begin{cases} 0, \quad& \text{if}\ 0\leq t< (a+b)/2-1/n\\
   1,\quad&\text{if}\ (a+b)/2+1/n\leq t\leq 1\\
   \text{a linear function},\ &\text{if}\ (a+b)/2-1/n\leq t<(a+b)/2+1/n
 \end{cases}
 \]
 is Cauchy but it is not convergent in $(X,\rho)$.

 In fact, $d(x_n, x_m) \leq \int_{a}^{b} (x_n(t) - x_m(t))
 \leq 4/\min\{n,m\}$ so that $\{x_n\}$ is Cauchy.

 If $x_n \rightarrow x$ as $n\to\infty$, then
 \[
 x(t) =
 \begin{cases}
   0 & \text{if } 0 \leq t \leq \frac{1}{2}, \\
   1 & \text{if } \frac{1}{2} < t \leq 1.
 \end{cases}
 \]
But this $x\not\in X$, which shows that $(X, \rho)$ (not $C[a,b]$)
is incomplete.
\end{proof}

\begin{Exec}
\label{Exec1.21}
  Suppose that $A$ is a nonempty subset of a metric space
  $(X,d)$.  Prove that the function $f:X\to\br$ defined by
  $f(x)=\inf\limits_{y\in A} d(x,y)$
  is continuous on $X$.  Furthermore $f$ is uniformly continuous on
  $X$.
\end{Exec}
\begin{proof}
For every $\varepsilon >0$, let $\delta = \varepsilon$.
 By the definition, there exists $y \in A$ such that
  $0 \leq d(x,y) - f(x) < \varepsilon $.
  So that $f(x') - f(x) \leq d(x',y) - f(x)
  \leq d(x',x) + d(x,y) - f(x) < 2 \varepsilon $
  for every $d(x,x') \leq \delta$. In the same way, we get that
  $f(x) - f(x') < 2 \varepsilon $ for every $d(x,x') \leq \delta$.
  Hence $|f(x) - f(x')| < 2 \varepsilon$, so that $f$ is uniformly continuous on $X$.
  \end{proof}

\begin{Exec}
\label{Exec1.22}
  Consider the metric space $\bc$ with the euclidean metric.  For every subset $A\subset\bc$ and every point
$z\in\bc$, let $\rho(z,A) = \inf_{ a\in A} |z - a|$.
\begin{itemize}
\item[d)] Prove that c) holds in an arbitrary metric space,
  i.e. if $A$ is a subset of a metric space $(X,d)$,
  then $x\in \overline{A}$ if and only if
  $$
  d(x,A)=\inf_{a\in A}d(x,a)=0.
  $$
\end{itemize}
\end{Exec}

\begin{proof}
  Suppose that $x\in \overline{A}$, then there exists a sequence
  $\{x_n\} \in A$ such that $x_n \rightarrow x$
  as $n \rightarrow 0$. Hence
  \[
  0 \leq \inf_{a\in A} d(x, a) \leq d(x, x_n) \rightarrow 0.
  \] so that $d(x,A) = 0$.
  On the other hand, if $d(x,A) = \inf_{a\in A}d(x,a)=0 $, there
  exists a sequence $\{x_n\} \in A$ such that
  $d(x,x_n) \rightarrow  \inf_{a\in A}d(x,a) = 0$ as $n \rightarrow \infty$.
  Thus $x \in \overline{A}$.
\end{proof}

\begin{Exec}\label{Exec1.23}
  Suppose that $(X,d)$ is a metric space, $F_1$ and $F_2$
  are closed subsets of $X$ with $F_1\cap\, F_2=\emptyset$.  Prove that
  there exists a continuous function on $X$ such that $f(x)=0$ if $x\in
  F_1$, and $f(x)=1$ if $x\in F_2$.
\end{Exec}

\begin{proof}
  Denote $d(x,A)=\inf_{y\in A}d(x,y)$ and let
  $$
  f(x)=\dfrac{d(x, F_1)}{d(x,F_1)+d(x,F_2)}.
  $$
  By Exercises~\ref{Exec1.21},  $f(x)$ is continuous.
  Since $F_1, F_2$ are closed, it follows that
  $d(x, F1) = 0$ and  $d(x, F2) = 0$  by Exercises~\ref{Exec1.22}. Hence $f(x)=0$ if $x \in
  F_1$ and $f(x)=1$ if $x \in F_2$.
\end{proof}

\begin{Exec}
  Suppose that $F_1$ and $F_2$ are as in
  Exercise~\ref{Exec1.23}.  Prove that there exist open sets $G_1$ and $G_2$ of $X$ such that
 $G_1\cap\, G_2=\emptyset$, $G_1\supset F_1$ and $G_2\supset F_2$. (Hint: Use the previous exercise and let
 $G_1=\{x\in X:f(x)<1/2\}$ and $G_2=\{x\in X: f(x)>1/2\}$.)
\end{Exec}
\begin{proof}
  Let $G_1=\{x\in X:f(x)<1/2\}$ and $G_2=\{x\in X: f(x)>1/2\}$.
  Clearly $F_1 \subset G_1$, $F_2 \subset G_2$ and $G_1\cap G_2 = \emptyset$
  follows from that $[0, 1/2) \cap (1/2, \infty)  = \emptyset$ and
  that
  $f$ is a function.
  $G_1, G2$ are open as an consequence of Theorem~\ref{Thm1.6.2} and
  Exercise~\ref{Exec1.21}.
\end{proof}

\begin{Exec}
  suppose that $(X,d)$ is a metric space, $f$ is a
  real function (mapping) on $X$.  Prove that $f$ is continuous on
  $X$ if and only if for each $r\in\br$, the sets $\{x\in X: f(x)\leq
  r\}$ and $\{x\in X; f(x)\geq r\}$ are all closed in $X$.
\end{Exec}
\begin{proof}
  Suppose that $f$ is continuous. Since sets $(-\infty,r]$ and $[r, \infty)$
  are closed, $\{x\in X: f(x)\leq r\}$ and $\{x\in X; f(x)\geq r\}$ are closed.

  Suppose that $x_0$ is a point of $X$.
  For every $\varepsilon > 0 $,
  since $\{x\in X: f(x)\geq f(x_0) + \varepsilon\}$
  is closed and $x_0$ not in it, there exists a real number $r_1 >0$
  such that $f(x) < f(x_0) + \varepsilon$ for every $x \in B(x_0,r_1)$.
  In the same way, there exists a number $r_2 >0$
  such that $f(x) > f(x_0) - \varepsilon$ for every $x \in B(x_0,r_2)$.
  Let $r = \min\{r_1, r_2\}$, then $|f(x) - f(x_0)| \leq \varepsilon$
  for every $x \in B(x_0,r)$.
  Note now that this argument is valid for every $x_0 \in X$.
  Hence $f$ is continuous at $x_0$ so that $f$ is continuous on $X$.
\end{proof}

\begin{Exec}
  Let $(X,d)$ and $(Y,\rho)$ be metric spaces and let $f:X\to
  Y$ be continuous.  Suppose that $A$ is  dense in $X$.
  Prove that $f(A)$ is dense in $f(X)$.
\end{Exec}

\begin{proof}
  Suppose that $y \in f(X)$, there exists a point $x \in X$
  such that $f(x) = y$.
  Since $A$ is dense in $X$, there exists a sequence $\{x_n\}$ in $A$
  such that $x_n \rightarrow x$ as $n \rightarrow \infty$.
  Clearly $\{f(x_n)\}$ is a sequence in $f(A)$.
  Hence $f(x_n) \rightarrow f(x) = y$ follows from that $f$ is continuous
  so that $f(A)$ is dense in $f(X)$.
\end{proof}

\begin{Exec}
  Denote $\br^\ast=\br\cup\{+\infty\}\cup\{-\infty\}$.
  Try to give a metric $\rho$ on $\br^\ast$ such that
  $(\br^\ast,\rho)$ and $[0.1]$ with the euclidean metric are
  isometric and homeomorphic.
\end{Exec}

\begin{solution}
  Consider
  $\rho(x,y)=\frac1\pi|\arctan x-\arctan y|$, the isometry
  $$
  f(x)=\begin{cases}
    \tan (\pi x-\pi/2),\quad&\text{if}\ x\in (0,1)\\
    -\infty,\quad&\text{if}\ x=0\\
    +\infty,\quad&\text{if}\ x=1.
  \end{cases}
  $$
\end{solution}

\begin{Exec}
  Denote $\tilde{\br}=\br\cup\{+\infty\}$. Try to give
  a metric $\rho$ on $\tilde{\br}$ such that $(\tilde{\br},\rho)$
  and $\{(x,y)\in\br^2: x^2+y^2=1\}$
  (i.e. the unit circle in $\br^2$) with the euclidean metric
  are isometric and homeomorphic.
\end{Exec}

\begin{solution}
  Consider
  $\rho(x,y) = \left( (\cos\arctan x - \cos\arctan y)^2
    + (\sin\arctan x - \sin\arctan y)^2
  \right)^{1/2}$,
  the isometry
  \[
  f(x)=\begin{cases}
    (\cos\arctan x, \sin\arctan x) & \text{if } x \in \br \\
    (0,1) & \text{if } x=\infty .
  \end{cases}
  \]
\end{solution}

\begin{Exec}
  Prove that a set in $\br^n$ is compact if and only if
  it is bounded and closed.
\end{Exec}
\begin{proof}
  Clearly compact set is bounded and closed.
  Suppos that $A$ is a bounded and closed set in $\br^n$.
  Let  $\{x_n\}$ is a sequence in $A$.
  By the Bolzano-Weierstrass Theorem, since
  $\{x_n\}$ is bounded, there exist
  a subsequence $\{x_{n_k}\}$ of $\{x_n\}$
  and a accumulation point $x_0$ such that
  $x_{n_k} \rightarrow x_0$ as $k \rightarrow \infty$.

  On the other hand $A$ is closed, so that $x_0 \in A$.
  Thus $A$ is compact.
\end{proof}

\begin{Exec}\label{Exec1.30}
  Suppose that $E$ is a nonempty compact set in a metric space $(X, d)$.
  Prove that there exist $x, y\in E$
  such that $d(x, y) = \sup_{u,v\in E} d(u, v)$.
\end{Exec}

\begin{proof}
  %Since $E$ is a compact set, $E$ is bounded so that
  %$\sup_{u,v\in E} d(u, v) < \infty$.
  By the definition of '$\sup$',
  there exists a sequence $\{(u_n, v_n)\}$ such that
  $d(u_n, v_n) \rightarrow \sup_{u,v\in E} d(u, v)$
  Since $E$ is a compact set, there exist a subsequence
  $\{u_{n_i}\}$ and a point $x \in E$
  such that $u_{n_i} \rightarrow x$ as $i \rightarrow \infty$.
  Then $\{v_{n_i}\}$ is a sequence in $E$, it follows that
  there exists a subsequence $\{v_{n_{i_j}}\}$ of $\{v_{n_i}\}$
  and $y \in E$
  such that $v_{n_{i_j}} \rightarrow y$ as $j \rightarrow \infty$.
  Hence $d(x,y) = \lim_{j \rightarrow \infty} d(u_{n_{i_j}},v_{n_{i_j}})
  = \sup_{u,v\in E} d(u, v)$.
\end{proof}

\begin{Exec}
  Suppose that $F_1$ and $F_2$ are compact subsets of a metric space
  $(X,d)$. Prove that there exist $x_0\in F_1$, $y_0\in F_2$ such that
  $d(F_1,F_2)=d(x_0,y_0)$, where $d(F_1,F_2)$ is the distance of sets
  $F_1$ and $F_2$, defined by $d(F_1,F_2)=\inf\{d(x,y):x\in F_1, y\in
  F_2\}$.
\end{Exec}

\begin{proof}
  There exist sequences $\{x_n\}$ and $\{y_n\}$ such that
  $d(x_n, y_n) \rightarrow \inf\{d(x,y): x \in F_1, y\in F_2\}$
  as $n \rightarrow \infty$.
  Since $F_1$ is compact, there exist a subsequence  $x_{n_i}$
  and a point $x_0 \in F_1$
  such that $x_{n_i} \rightarrow x_0$ as $i \rightarrow \infty$.
 Since also $F_2$ is compact,
  there exist a subsequence  $y_{n_{i_j}}$
  and a point $y_0 \in F_2$
  such that $y_{n_{i_j}} \rightarrow y_0$ as $j \rightarrow \infty$.
  Hence $d(x_0,y_0) = d(F_1, F_2)$.
\end{proof}

\begin{Exec}
  Suppose that $F_1$ and $F_2$ are subsets of a metric space $(X,d)$,
  where $F_1$ is compact in $X$ and $F_2$ is closed in
  $X$.  Prove that if $d(F_1,F_2)=0$ then $F_1\cap
  \,F_2\not=\emptyset$.
\end{Exec}

\begin{proof}
  Since
  \[
  d(F_1,F_2) = \inf\{d(x,y):x\in F_1, y\in F_2\}
  = \inf_{x \in F_1} \inf_{y \in F_2} d(x,y) = \inf_{x \in F_1} d(x,F_2),
  \]
  there exists a sequence $\{x_n\}$
  such that $d(x_n, F_2) \rightarrow d(F_1, F_2)$ as $n \rightarrow \infty$.
  It follows by the compaction of $F_1$ that
  there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that
  $ x_{n_k} \rightarrow x$ as $k \rightarrow \infty$,
  where $x$ is a point in $F_1$.
  Suppose that $d(F_1, F_2) = 0$.
  Clearly $d(x, F_2) = \lim_{k \rightarrow \infty} d(x_{n_k}, F_2) = d(F_1, F_2) = 0$.
The closeness of  $F_2$  implies that $x\in F_2$ (by Exercises
\ref{Exec1.22}),
  so that $F_1 \cap F_2 \supset \{x\} \neq \emptyset$.
\end{proof}

\begin{Exec}
  Let $n$ be fixed and define $\rho(x,y)=\max|x_i-y_i|$ on $\br^n$.
  Let $y=Tx$ be defined by
  $$
  y_i=\sum_{j=1}^n a_{ij}x_j+b_i\quad ( 1\leq i\leq n ).
  $$
  Prove that $T$ is a contraction mapping on $(\br^n,\rho)$ if and
  only if
  $$
  \max_{i}\sum_{j=1}^n|a_{ij}|<1 .
  $$
\end{Exec}

\begin{proof}
  Suppose that $c = \max_{0 \leq i \leq n}\sum_{j=1}^n|a_{ij}|<1$.
  For every $x=(x_1, \dots, x_n)^T, y=(y_1, \dots, y_n)^T \in \br^n$,
  since
  \[
  \begin{split}
    \rho(Tx,Ty) &=
    \max_{0 \leq i \leq n} \left|
      \sum_{j=1}^{n} a_{ij} x_j + b_i
      - \sum_{j=1}^{n} a_{ij} y_j - b_i
    \right|
    =  \leq \max_{0 \leq i \leq n} \left|
      \sum_{j=1}^{n} |a_{ij}| (x_j - y_j)
    \right| \\
    &\leq \max_{0 \leq i \leq n}
    \sum_{j=1}^{n} |a_{ij}| |x_j - y_j|
    \leq \max_{0 \leq i \leq n} \sum_{j=1}^{n} |a_{ij}|
    \max_{0\leq j\leq n} |x_j - y_j|
    \leq c\; \rho(x,y),
  \end{split}
  \]
  which gives that $T$ is a contraction mapping.

  On the other hand, suppose that $T$ is contraction mapping.
  So there exists $c < 1$ such that $\rho(Tx, Ty) < c\rho(x, y)$ for
  every $x,y \in \br^n$.
  Let $y = (0, \dots, 0)^T$ and,
  for every $0 \leq i \leq n$,
  let $x^i = (x_1^i, \dots, x_n^i)^T$, where $x_j^i = 1$ if $a_{ij} \geq 0$
  and $x_j^i = -1$ if $a_{ij} <0$, then
  \[
  \sum_{j=1}^{n} |a_{ij}| = \sum_{j=1}^{n} a_{ij} (x_j^i-0)
  = \rho(Tx^i, y) \leq c \rho(x,y) = c.
  \]
  Thus $\max_{i}\sum_{j=1}^n|a_{ij}|<1$.
\end{proof}
\
\begin{Exec}
  Suppose that $(X,d)$ is complete, $T:X\to X$. Prove that if
  $$
  \alpha_0=\inf\limits_{n}\sup\limits_{x\not=
    y}\dfrac{d(T^nx,T^ny)}{d(x,y)}<1,
  $$
  then $T$ has a unique fixed point on $X$.
\end{Exec}

\begin{proof}
  Let $\varepsilon = (1-\alpha_0)/2$, there exists $N > 0$
  such that
  \[
  \sup\limits_{x\not=
    y}\dfrac{d(T^N x,T^N y)}{d(x,y)} - \alpha_0 < \varepsilon
  \]
  so that
  \[
  \sup\limits_{x\not=y}\dfrac{d(T^N x,T^N y)}{d(x,y)}
  \leq (1+\alpha_0)/2 \triangleq c <1.
  \]
  Clearly $d(T^N x, T^N y) \leq
  \sup\limits_{x\not=y}\dfrac{d(T^N x,T^N y)}{d(x,y)} d(x,y) = c\, d(x,y)$
  for every $x, y \in X$.
  By Theorem \ref{Thm1.8.2}, we get that $T$ has a unique fixed point on $X$.
\end{proof}

\begin{Exec}
  Suppose that $(X,d)$ is complete, the mapping $T:X\to X$
  satisfies $d(Tx,Ty)\leq \theta d(x,y)$ for all $x,y$ in the closed ball
  $S(x_0,r)$ and $d(x_0, Tx_0)<(1-\theta)r$, where $0\leq
  \theta<1$.  Prove that $T$ has a unique fixed point on
  $S(x_0,r)$.
\end{Exec}

\begin{proof}
  Since $(X, d)$ is complete and $S(x_0,r)$ is closed,
  $S(x_0,r)$ is a complete subspace of $(X,d)$.
  $T\,S(x_0,r) \subset S(x_0,r)$ follows from $d(x_0, Tx_0)<(1-\theta)r$.
  So $T$ is a contraction of $S(x_0,r)$.
  By Theorem~\ref{Thm1.8.1}, $T$ has a unique fixed point on $S(x_0,r)$.
\end{proof}

\begin{Exec}
  a) Can the completeness of $(X,d)$ in
  Theorem~\ref{Thm1.8.1} be removed?

  b) If contraction condition for $T$ is replaced by
  $d(Tx,Ty)<d(x,y)$ for all $x,y\in X$ and $x\not=y$,
  does Theorem~\ref{Thm1.8.1} hold?
\end{Exec}

\begin{proof}
  \begin{itemize}
  \item[a)] No. Consider that $X = (0,1)$, $d$ is the euclidean matric
    and  $f(x) = x/2$. Clearly $0 \not\in X$ and
    $0$ is the unique fixed point of $f$
    in $\br \supset X$ so that $f$ dose not have any fixed point in $X$.
  \item[b)] No. Consider that $X = [2,\infty)$, $d$ is the euclidean matric
    and $f(x) = x + 1/x$. Then $f(X) \subset X$ and
    $|f(x)-f(y)| = |(x-y) + (1/x - 1/y)| = |x-y||1- 1/xy| < |x-y|$
    for every $x,y \in X$. Clearly $f$ dose not have any fixed point in $X$.
  \end{itemize}
\end{proof}

\begin{Exec}
   Let $(X,d)$ be compact.  Suppose that a mapping $T:X\to
   X$ satisfies that $d(Tx,Ty)<d(x,y)$ for all $x,y\in X$ and
   $x\not=y$.  Prove that $T$ has a unique fixed point on $X$.
\end{Exec}

\begin{proof}
  Let
  \[
  c = \sup_{x \neq y} \frac{d(Tx,Ty)}{d(x,y)},
  \]
  then $c \leq 1$ since $d(Tx,Ty)<d(x,y)$ for all $x,y\in X$($x\neq y$).
  Thus there exist sequences $\{x_n\}, \{y_n\}\subset X$ such that
  $d(Tx_n,Ty_n)/d(x_n,y_n) \rightarrow c$.
  In the same way as Execrises~\ref{Exec1.30}, Since $X$ is compact,
  we get there exists a sequence $\{n_k\}$ such that
  $x_{n_k} \rightarrow x_0$ and $y_{n_k} \rightarrow y_0$, where $x_0, y_0 \in X$.
  So $c = d(Tx_0,Ty_0)/d(x_0,y_0) < 1$. Hence $T$ is a contraction.
  By Theorem~\ref{Thm1.8.1}, $T$ has a unique fixed point.
\end{proof}

\begin{Exec}
  Let $f(t)\in C[0,1]$. Find the continuous solution
  $x(t)$ of the integral equation
  $$
  x(t)=f(t)+\lambda\int_0^t x(s)\,
  ds\ \quad (t\in [0,1]).
  $$
\end{Exec}

\begin{solution}
  Let $T = f(t)+\lambda\int_0^t x(s)\, ds\ \quad (t\in [0,1]).$
  By Example~\ref{Exam1.8.4}, let $K(t,s)=1$, we get $T$
  has a unique fixed point
  in $C[0,1]$. So the solution of the integral equation
  $x = \lim_{n\rightarrow \infty} T^n 0$.
  Hence
  \[
  \aligned
  x(t) =& f(t) + \lambda \int_0^t f(s_0)ds_0 +
  \lambda^2 \int_0^t \int_0^{s_0} f(s_1) ds_1 ds_0 + \cdots \\
  &\quad +
  \lambda^n \int_0^t \dots \int_0^{s_n} f(s_n) ds_n\dots ds_0 + \cdots.
  \endaligned
  \]
\end{solution}

\begin{Exec}
  Suppose that $(a_{ij}), ij=1,2,\cdots$, is an
    infinite matrix.  Let
    $$
\sup_{i\in\bn}\sum_{j=1}^\infty |a_{ij}|<1.
    $$
    Prove that the system of equations
    $$
x_i=\sum_{j=1}^\infty a_{ij}x_j+b_i\quad (i=1,2,\cdots),
    $$
    where $(b_i)$ is bounded, has a unique bounded solution.
\end{Exec}

\begin{proof}
  Suppose that $x,y \in l^\infty$.
  Let $Tx = \{ \sum_{j=1}^\infty a_{ij}x_j+b_i \}$ and
  $c = \sup_{i\in\bn}\sum_{j=1}^\infty |a_{ij}|$, then $c<1$.
  Since $\tilde{x}_i = \sum_{j=1}^\infty a_{ij}x_j+b_i
  \leq c \sup_{j\in\bn} x_j  + \sup_{i\in\bn} b_i < \infty$,
  it follows that $Tl^\infty \subset l^\infty$.
  Clearly $T$ is a contraction since
  \[
  \begin{split}
    d(Tx,Ty) &= \sup_{i\in\bn}\left|
      \sum_{j=1}^\infty a_{ij}x_j+b_i - \sum_{j=1}^\infty a_{ij}y_j+b_i
    \right| \\
    &=  \sup_{i\in\bn}\left| \sum_{j=1}^\infty a_{ij} (x_j-y_j) \right|
    \leq \sup_{i\in\bn}\left| \sum_{j=1}^\infty a_{ij} \right| \sup_{j \in \bn} |x_j-y_j|
    = c d(x,y).
  \end{split}
  \]
  By Theorem~\ref{Thm1.8.1} $T$ has a unique bounded fixed point so that
  the equations has a unique bounded solution.
\end{proof}

\docend
